I have this question I'm unable to solve (you need to prove this or give a counterexample:
If a sequence $f_{n}\neq0$ ($f_{n}\left(x\right)\neq0,\forall n\geq N\in\mathbb{N}$ for some N) converges uniformly and $\frac{1}{f_{n}}$ converges pointwise then it also converges uniformly.
I tried proving $$\lim_{n\to\infty}\sup\left|\frac{1}{f_{n}\left(x\right)}-\frac{1}{f\left(x\right)}\right|=0$$ but it didn't work. I tried looking at function of the form $\frac{1}{n}$ but $\frac{1}{\frac{1}{n}}=n$ doesn't converge. I tried looking at $$f_{n}\left(x\right)=\begin{cases} \frac{1}{n} & x\leq\frac{1}{n}\\ 0 & x>\frac{1}{n} \end{cases}$$ but it doesn't fit the demand of $f_{n}\neq0$.
So I wasn't able to prove nor provide a counterexample
A counterexample is $f_n(x) = \frac{1}{x} + \frac{1}{n}$ where $f_n(x) \to f(x) =\frac{1}{x}$ uniformly on $(0,\infty)$ and $\frac{1}{f_n(x)} \to x$ pointwise.
The uniform convergence of $f_n$ is obvious since $\left|f_n(x) - f(x)\right| = \frac{1}{n}$. Choosing any $N(\epsilon) > \frac{1}{\epsilon}$ we have $|f_n(x) - f(x)| <\frac{1}{n} < \frac{1}{N(\epsilon)}< \epsilon$ for all $n > N(\epsilon)$ and $x \in (0,\infty)$.
However,
$$\left|\frac{1}{f_n(x)}- \frac{1}{f(x)} \right|= \left|\frac{1}{\frac{1}{x} + \frac{1}{n}}- x\right|= \frac{\frac{x}{n}}{\frac{1}{x} + \frac{1}{n}} =\frac{x^2}{n + x},$$
and the convergence of $\frac{1}{f_n}$ is not uniform on $(0,\infty)$ since
$$\sup_{x \in (0,\infty)}\left|\frac{1}{f_n(x)}- \frac{1}{f(x)} \right| = \sup_{x \in (0,\infty)}\frac{x^2}{n + x}\geqslant \frac{n^2}{n+ n}= \frac{n}{2}\underset{n \to \infty}\longrightarrow+\infty \neq 0 $$